In the last post we proved that $F_1\Ccl$ is a category. Consider now the category $F_1^+\Ccl$ whose objects are pairs $(A\to A', A'/A)$ where $A\to A'$ is an object of $F_1\Ccl$ and $A'/A$ is a choice of pushout of $*\leftarrow A\to A'$. In other words, we can think of $F_1^+\Ccl$ to be the category of sequences $A\to A'\to A'/A$ where $A'/A$ is the pushout. Perhaps even better, we can think of $F_1^+\Ccl$ as the category whose objects are all commutative squares

that are pushout squares. Here I used the symbol $B$ to denote an arbitrary pushout as there may be more than one choice of pushout. In general I'll just write $A'/A$ as I'll just be talking about one choice at a time.

Notice that a morphism $(A\to A')\to (B\to B')$ in $F_1\Ccl$ determines a morphism of pushout squares an vice-versa: indeed, given such a morphism, there is a unique morphism $A'/A\to B'/B$. Hence, the functor $F_1^+\Ccl\to F_1\Ccl$ given by sending $A\to A'\to A'/A$ to $A\to A'$ is an equivalence of categories because it is fully faithful and essentially surjective. This equivalence naturally gives $F_1^+\Ccl$ the structure of a category with cofibrations.

There are also functors $s,t,q:F_1^+\Ccl\to\Ccl$ given by:

$s(A\to A'\to A'/A) = A$

$t(A\to A'\to A'/A) = A'$

$q(A\to A'\to A'/A) = A'/A$

We will now see in the rest of the post that these three functors $s,t,q$ are exact functors; in other words, they preserve the structure of a category with cofibrations. In the following proofs, we won't say anything about the zero object, since that part is trivial. Also, that $s$ is exact is pretty straightforward, but it will be a good warmup for the harder ones.

Theorem. The functor $s:F_1^+\Ccl\to \Ccl$ given by $s(A\to A'\to A'/A) = A$ is an exact functor.

Proof. We first need to show that $s$ preservs cofibrations. So suppose that $f:(A\to A'\to A'/A)\to (B\to B'\to B'/B)$ is a cofibration. By definition, $A\to B$ is a cofibration and $s(f) = A\to B$.

Next, let us show that $s$ preserves the pushout square in $F_1^+\Ccl$. What is the pushout square? Once we choose a pushout for the diagram $C'/C\leftarrow A'/A\to B'/B$, say $(B'/B)\cup_{A'/A}C'/C$ we get a pushout square in $F_1^+\Ccl$:
You should convince yourself that this is actually a pushout in $F_1^+\Ccl$. Of course, we didn't really need to know the entire pushout square to see that applying $s$ to the diagram gives the familiar pushout square with pushout $B\cup_A C$.

Theorem. The functor $t:F_1^+\Ccl\to \Ccl$ given by $s(A\to A'\to A'/A) = A'$ is an exact functor.

Proof. Referring to the big diagram above, applying $t$ to a cofibration $(A\to A'\to A'/A)\to (B\to B'\to B'/B)$ gives $A'\to B'$. We have already remarked that this must be a cofibration in $\Ccl$, but let's go through the argument again: by definition, we know that $A'\cup_AB\to B'$ is a cofibration in $\Ccl$. And, $A'\to A'\cup_AB$ is the pushout along $A\to B$, which is a cofibration in $\Ccl$, so $A'\to A'\cup_AB$ is also a cofibration in $\Ccl$. Hence the composition $A'\to A'\cup_AB\to B'=A'\to B'$ is a cofibration in $\Ccl$.

Now, we need to show that $t$ takes the big pushout square to a pushout square that satisfies the third axiom of a category with cofibrations. From the way the big pushout was constructed, this is again clear.

Finally, the nontrivial and tricky part!

Theorem. The functor $s:F_1^+\Ccl\to \Ccl$ given by $s(A\to A'\to A'/A) = A'/A$ is an exact functor.

Proof. Suppose $(A\to A'\to A'/A)\to (B\to B'\to B'/B)$ is a cofibration in $F_1^+\Ccl$; in other words, $(A\to A')\to (B\to B')$ is a cofibration in $F_1\Ccl$. We need to show that the natural map $A'/A\to B'/B$ is a cofibration back in $\Ccl$. This follows from considering three pushout squares:
The three inner squares are pushouts, so the two big rectangles give the isomorphisms and the required map, being a pushout along $A'\cup_AB\to B'$. (One should compare this proof to the diagram-chasing proof for abelian groups and cofibrations=injections.)

Now we have to show that $q$ takes pushouts to pushouts. Let us again recall the big pushout diagram

Now we already mentioned that this is actually a pushout square, though we didn't prove it. This comes down to proving that $B'/B\cup_{A'/A}C'/C$ is actually a choice for the quotient $(B'\cup_{A'}C')/(B\cup_AC)$. To do so, first assume we have a commutative diagram

We would like to show that there exists a unique arrow $(B'\cup_A' C')/(B\cup_A C)\to X$ making the entire diagram commute. Do do so, first note that this diagram gives a commutative diagram

which in turn, together with the first diagram gives the commutative diagram
Which gives the desired arrow (I've left out a few steps, here, but these are straightforward diagram manipulations. Check them!)

In a few more posts we will be coming to the point where we can define a Waldhausen category, or a category with cofibrations and weak equivalences. It's probably helpful to look at a rough roadmap of our immediate goals:

Define a Waldhausen category

Define the $K$-theory of a Waldhausen category

Show that this agrees with the "obvious" definition for $K_0$.

State the additivity theorem and explain its importance

Prove the additivity theorem

Deduce various corollaries from the additivity theorem

After these goals, I shall probably go through the rest of Section 1, including explaining how $K$-theory from Waldhausen's construction gives the same thing as Quillen's Q-construction for exact categories.